Table of Laplace transforms

The error function erf is defined by

1. Linearity: \( {\cal L} \left[ \alpha\,f(t) + \beta\, g(t) \right] = \alpha\, {\cal L} \left[ f \right] + \beta\,{\cal L} \left[ g \right] = \alpha\, f^L + \beta \, g^L . \) Proof:
   For arbitrary cosnatnts α and β, we have

   \begin{align*} {\cal L} \left[ \alpha f(t) + \beta g(t) \right] &= \int_0^{\infty} e^{-\lambda t} \left[ \alpha f(t) + \beta g(t) \right] {\text d}t \\ &= \int_0^{\infty}\left[ \alpha f(t)\, e^{-\lambda t} + \beta g(t)\, e^{-\lambda t} \right] {\text d}t \\ &= \int_0^{\infty}\alpha f(t)\, e^{-\lambda t} {\text d}t + \int_0^{\infty} \beta g(t) \, e^{-\lambda t} {\text d}t \\ &= \alpha \int_0^{\infty} f(t)\, e^{-\lambda t} {\text d}t + \beta \int_0^{\infty} g(t) \, e^{-\lambda t} {\text d}t \\ &= \alpha {\cal L} \left[ f(t) \right] + \beta {\cal L} \left[ g(t) \right] = \alpha\,f^L + \beta \,g^L . \end{align*}
2. The derivative rule:
   \begin{equation} \label{EqTable.2} {\cal L} \left[ f^{(n)} (t) \right] = \lambda^n - \sum_{k=1}^n \lambda^{n-k} f^{(k-1)} (+0) . \end{equation}
   In particular,
   \begin{equation} \label{EqTable.3} {\cal L} \left[ f' (t) \right] = {\cal L} \left[ \frac{{\text d}f (t)}{{\text d}t} \right] = \lambda\,f^L (\lambda ) - f(+0) . \end{equation}
   \begin{equation} \label{EqTable.4} {\cal L} \left[ f'' (t) \right] = {\cal L} \left[ \frac{{\text d}^2 f (t)}{{\text d}t^2} \right] = \lambda^2 f^L (\lambda ) - f'(+0) - \lambda\,f(+0) . \end{equation}
   Proof:
   By mathematical induction, utilizing integration by parts

   \[ {\cal L} \left[ f' (t) \right] = \int_0^{\infty} e^{-\lambda t} f' (t)\, {\text d}t = \left. e^{-\lambda t} f (t) \right\vert_{t=0}^{t=\infty} - \lambda \int_0^{\infty} e^{-\lambda t} f (t)\, {\text d}t = -f(+0) + \lambda\,f^L (\lambda ) . \] Assuming the property holds for n = k ∈ ℤ⁺, we have

   \[ {\cal L} \left[ f^{(k)} (t) \right] = \lambda^k f^L - \lambda^{k-1} f(+0) - \lambda^{k-2} f' (+0) - \cdots - \lambda\,f^{(k-2)} (+0) - f^{(k-1)} (+0) . \]

   We now prove for n = k + 1.

   \begin{align*} {\cal L} \left[ f^{(k+1)}(t) \right] &= \int_0^{\infty} e^{-\lambda t} f^{(k+1)}(t) \,{\text d}t = \lim_{A\to \infty} \int_0^A e^{-\lambda t} f^{(k+1)}(t) \,{\text d}t \\ &= \lim_{A\to \infty} \left[ \left. e^{-\lambda t} f^{(k)}(t) \right\vert_{t=0}^{t=A} + \lambda \int_0^A f^{(k)}(t) e^{-\lambda t} {\text d}t \right] \\ &= \lambda\,{\cal L} \left[ f^{(k)}(t) \right] - f^{(k)}(+0) \end{align*}

   and the statement follows.
3. Convolution rule: \( {\cal L} \left[ f \ast g \right] = f^L \,g^L . \)
   Recall that the convolution of two functions f and g is
   \begin{equation} \label{EqTable.1} \left( f \ast g \right) (t) = \int_0^t f(t-\tau )\,g(\tau ) \,{\text d} \tau = \int_0^t g(t-\tau )\,f(\tau ) \,{\text d} \tau = \left( g \ast f\right) (t) . \end{equation}
   Proof:
   Application of the Laplace transformation to the convolution integral \eqref{EqTable.1} yields

   \[ {\cal L} \left[ f*g \right] (\lambda ) = \int_0^{\infty} e^{-\lambda t} {\text d} t \int_0^{t} f(\tau )\, g(t-\tau )\, {\text d}\tau . \]

   Reversing the order of integration gives

   \[ {\cal L} \left[ f*g \right] (\lambda ) = \int_0^{\infty} {\text d}\tau .f(\tau ) \, e^{-\lambda \tau} \int_{\tau}^{\infty} {\text d} t\,g(t-\tau )\, e^{-\lambda (t- \tau )} \]

   Substitution u = −τ shows that

   \[ {\cal L} \left[ f*g \right] (\lambda ) = \int_0^{\infty} {\text d}\tau .f(\tau ) \, e^{-\lambda \tau} \int_{0}^{\infty} {\text d} u\, g(u) \, e^{-\lambda u} = f^L \cdot g^L . \]

   To finish the proof, we need to show that the convolution is a function-original.
4. Shift rule: \( {\cal L} \left[ f(t-a)\,H(t-a) \right] = e^{-a\lambda} \,f^L (\lambda ) . \) Proof:
   \begin{align*} {\cal L} \left[ H(t-a)\,f(t-a) \right] &= \int_0^{\infty} e^{-\lambda t} H(t-a)\,f(t-a)\, {\text d} t = \int_a^{\infty} e^{-\lambda t}\,f(t-a)\, {\text d} t \\ &= \int_0^{\infty} e^{-\lambda \left( \tau + a \right)} f\left( \tau \right) {\text d} \tau \end{align*}

   because of substitution t = τ + 𝑎. Then

   \[ {\cal L} \left[ H(t-a)\,f(t-a) \right] = \int_0^{\infty} e^{-\lambda \left( \tau + a \right)} f\left( \tau \right) {\text d} \tau = e^{-\lambda a} f^L (\lambda ) . \]
5. Similarity rule: \( {\cal L} \left[ f(kt) \right] = \frac{1}{k}\, f^L \left( \frac{\lambda}{k} \right) . \) Proof:
   \begin{align*} {\cal L} \left[ f(kt) \right] &= \int_0^{\infty} e^{-\lambda t} f(kt)\,{\text d}t = \frac{1}{k} \int_0^{\infty} e^{-\lambda u/k} f(u)\,{\text d}u \\ &= \frac{1}{k} \, f^L \left( \frac{\lambda}{k} \right) . \end{align*}
6. Attenuation rule: \( {\cal L} \left[ e^{-at} \, f(t) \right] = f^L \left( \lambda +a \right) . \) Proof:
   \[ {\cal L} \left[ e^{-at} \, f(t) \right] = \int_0^{\infty} e^{-at} e^{-\lambda t} f(t)\, {\text d}t = \int_0^{\infty} e^{-at - \lambda t} f(t)\, {\text d}t = f^L (\lambda + a ) . \]
7. Differentiation rule: \( \frac{{\text d}}{{\text d} \lambda} \, f^L (\lambda ) = - {\cal L} \left[ t\, f(t) \right] . \) In deneral, for a positive integer n,
   \[ {\cal L} \left[ t^n f(t) \right] = \left( -1 \right)^n \frac{{\text d}^n}{{\text d}\lambda^n} \,{\cal L} \left[ f(t) \right] = \left( -1 \right)^n \frac{{\text d}^n}{{\text d}\lambda^n} \,f^L (\lambda ) . \]
   Proof:
   Using mathematical induction,

   \[ {\cal L} \left[ t\, f(t) \right] = \int_0^{\infty} t\,f(t)\,e^{-\lambda t} {\text d}t = - \int_0^{\infty} f(t)\,\frac{\text d}{{\text d}\lambda}\,e^{-\lambda t} {\text d}t = -\frac{\text d}{{\text d}\lambda} \int_0^{\infty} f(t)\,e^{-\lambda t} {\text d}t = - \frac{\text d}{{\text d}\lambda} \,f^L (\lambda ) . \]

   Assuming that the property holds for n = k ∈ ℤ₊, then

   \[ {\cal L} \left[ t^k f(t) \right] = \left( -1 \right)^k \frac{{\text d}^k}{{\text d}\lambda^k} \,{\cal L} \left[ f(t) \right] = \left( -1 \right)^k \frac{{\text d}^n}{{\text d}\lambda^k} \,f^L (\lambda ) . \]

   We now prove that the property holds for n = k+1,

   \begin{align*} \left( -1 \right)^{k+1} \frac{{\text d}^{k+1}}{{\text d}\lambda^{k+1}}\, f^L (\lambda ) &= \left( -1 \right) \left( -1 \right)^k \frac{{\text d}}{{\text d}\lambda} \left[ \frac{{\text d}^{k}}{{\text d}\lambda^{k}}\, f^L (\lambda ) \right] \\ &\quad \\ &= \left( -1 \right) \frac{{\text d}}{{\text d}\lambda} \left[ \left( -1 \right)^k \frac{{\text d}^{k}}{{\text d}\lambda^{k}}\, f^L (\lambda ) \right] = \left( -1 \right) \frac{{\text d}}{{\text d}\lambda} \,{\cal L} \left[ t^k f(t) \right] \\ &\quad \\ &= \left( -1 \right) \frac{{\text d}}{{\text d}\lambda} \int_0^{\infty} \left[ t^k f(t) \, e^{-\lambda t} \right] {\text d} t \\ &\quad \\ &= \int_0^{\infty} \frac{\partial}{\partial \lambda} \left[ t^k f(t)\, e^{-\lambda t} \right] {\text d} t \\ &\quad \\ &= \int_0^{\infty} e^{-\lambda t} t^{k+1} f(t) \, {\text d} t = {\cal L} \left[ t^{k+1} f(t) \right] . \end{align*}
8. Integration rule: \( {\cal L} \left[ t^n \ast f(t) \right] = \frac{n!}{\lambda^{n+1}} \, f^L (\lambda ) . \)
   In particular, for n = 0,
   \[ {\cal L} \left[ \int_0^t f(s)\,{\text d}s \right] = \frac{1}{\lambda}\, f^L (\lambda ) . \]
   Proof:
   The property follows from the convolution rule.
9. The Laplace transform of periodic functions.
   If \( f(t) = f(t+ \omega ) , \) then \( \displaystyle f^L (\lambda ) = \int_0^{\infty} f(t )\,e^{-\lambda\,t} \,{\text d} t = \frac{1}{1- e^{-\omega\lambda}} \, \int_0^{\omega} \,f(t ) \,e^{-\lambda \, t} \,{\text d} t . \) Proof:
   If f(t) is periodic with period ω and its Laplace transform exists, then

   \begin{align*} f^L (\lambda ) &= {\cal L} \left[ f(t) \right] = \int_0^{\infty} e^{-\lambda t} f(t)\, {\text d}t \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + \int_{\omega}^{\infty} e^{-\lambda t} f(t)\, {\text d}t \qquad \mbox{make substitution at the latter} \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + \int_0^{\infty} e^{-\lambda (u+\omega )} f(u + \omega )\, {\text d}u \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + e^{-\lambda\omega} \int_0^{\infty} e^{-\lambda u} f(u)\, {\text d}u \\ &= \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + e^{-\lambda\omega} f^L . \end{align*}

   The required identity follows from the equation

   \[ f^L = \int_0^{\omega} e^{-\lambda t} f(t)\, {\text d}t + e^{-\lambda\omega} f^L . \]
10. The Laplace transform of anti-periodic functions.
    If \( f(t) = -f(t+ \omega ) , \) then \( \displaystyle f^L (\lambda ) = \int_0^{\infty} f(t )\,e^{-\lambda\,t} \,{\text d} t = \frac{1}{1+ e^{-\omega\lambda}} \, \int_0^{\omega} \,f(t ) \,e^{-\lambda \, t} \,{\text d} t . \) Proof:
    Proof is similar to the previous one.
11. Division by t:
    \( \displaystyle {\cal L} \left[ \frac{f(t)}{t} \right] (\lambda ) = \int_{\lambda}^{\infty} f^L (u)\,{\text d}u . \) Proof:
    \begin{align*} \int_{\lambda}^{\infty} f^L (s)\,{\text d}s &= \int_{\lambda}^{\infty} {\text d}s \int_0^{\infty} e^{-st} f(t) \,{\text d}t \\ &= \int_0^{\infty} {\text d}t \int_{\lambda}^{\infty} e^{-st} f (s)\,{\text d}s \\ &= \int_0^{\infty} {\text d}t\left[ \lim_{k\to\infty} \int_{\lambda}^{k} e^{-st} f (s)\,{\text d}s \right] \\ &= \int_0^{\infty} {\text d}t\left[ \lim_{k\to\infty} \left. f(t) \,\frac{e^{-st}}{-t} \right\vert_{s}^k \right] = \int_0^{\infty} {\text d}t \lim_{k\to\infty} \left[ e^{-st} - e^{-kt} \right] \frac{f(t)}{t} \\ &= {\cal L} \left[ \frac{1}{t}\, f(t) \right] . \end{align*}

Table of Laplace transform properties in Mathematica:

Example 1: Consider a linear function f(t) = t on the interval [0, ∞). For a positive T, its full-wave rectifier is

\[ g(t) = t \left[ H(t) - H(t-T) \right] + \left( t-T \right) \left[ H(t-T) - H(t-2T) \right] + \left( t-2T \right) \left[ H(t-2T) - H(t-3T) \right] + \cdots . \]

		We plot full wave rectifier of function f(t) = t: f[t_] = Piecewise[{{t, 0 < t < 1}, {t-1, 1 < t < 2}, {t - 2, 2 < t < 3}, {t-3, 3 < t < 4}, {t - 4, 4 < t < 5}}]; plot = Plot[f[t], {t, 0, 3.5}, PlotStyle -> Thickness[0.01], AspectRatio -> 1/3, Axes -> False, PlotRange -> {{-0.4, 3.8}, {-0.6, 1.5}}]; ar = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{-0.2, 0}, {3.8, 0}}]}]; ar2 = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{0, -0.2}, {0, 1.2}}]}]; tt = Graphics[{Black, Text[Style["t", 18], {3.6, 0.3}]}]; t1 = Graphics[{Black, Text[Style["T", 18], {1, -0.3}]}]; t2 = Graphics[{Black, Text[Style["2T", 18], {2, -0.3}]}]; t3 = Graphics[{Black, Text[Style["3T", 18], {3, -0.3}]}]; Show[plot, ar, ar2, tt, t1, t2, t3]
Full-wave rectifier of function f = t.		Mathematica code

Since the full wave rectifier is a periodic function with period T, its Laplace transform is

\[ {\cal L} \left[ g(t) \right] = \frac{1}{1 - e^{-\lambda T}} \, \int_0^T t\,e^{-\lambda t} {\text d}t = \frac{1}{\lambda^2} - \frac{T}{1 - e^{-\lambda T}} \,\frac{1}{\lambda} \,e^{-\lambda T} . \]

Assuming[ ss > 0, Integrate[t*Exp[-ss*t], {t, 0, TT}]]

(1 - E^(-ss TT) (1 + ss TT))/ss^2

 

Now we consider the half-wave rectifier that corresponds to the linear function. It is a 2T periodic function:

\[ h(t) = t \left[ H(t) - H(t-T) \right] + \left( t-2T \right) \left[ H(t-2T) - H(t-3T) \right] + \left( t-4T \right) \left[ H(t-4T) - H(t-5T) \right] + \cdots . \]

		We plot full wave rectifier of function f(t) = t: f[t_] = Piecewise[{{t, 0 < t < 1}, {0, 1 < t < 2}, {t - 2, 2 < t < 3}, {0, 3 < t < 4}, {t - 4, 4 < t < 5}}]; plot = Plot[f[t], {t, 0, 5.3}, PlotStyle -> Thickness[0.01], AspectRatio -> 1/3, Axes -> False, PlotRange -> {{-0.4, 5.5}, {-0.6, 1.5}}]; ar = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{-0.2, 0}, {5.5, 0}}]}]; ar2 = Graphics[{Black, Thickness[0.01], Arrowheads[0.08], Arrow[{{0, -0.2}, {0, 1.2}}]}]; tt = Graphics[{Black, Text[Style["t", 18], {5.2, 0.3}]}]; t1 = Graphics[{Black, Text[Style["T", 18], {1, -0.3}]}]; t2 = Graphics[{Black, Text[Style["2T", 18], {2, -0.3}]}]; t3 = Graphics[{Black, Text[Style["3T", 18], {3, -0.3}]}]; t4 = Graphics[{Black, Text[Style["4T", 18], {4, -0.3}]}]; Show[plot, ar, ar2, tt, t1, t2, t3, t4]
Half-wave rectifier of function f = t.		Mathematica code

Again, using rule 9, we obtain the Laplace transform of the half-wave rectifier:

\[ {\cal L} \left[ h(t) \right] = \frac{1}{1 - e^{-2\lambda T}} \, \int_0^T t\,e^{-\lambda t} {\text d}t = \frac{1}{1 - e^{-2\lambda T}} \,\frac{1}{\lambda^2} \left( 1 - e^{-\lambda T} \right) \left( 1 + \lambda T \right) = \frac{1}{1 + e^{-\lambda T}} \,\frac{1}{\lambda^2} \left( 1 + \lambda T \right) . \]

   ■

Return to Mathematica page
Return to the main page (APMA0330)
Return to the Part 1 (Plotting)
Return to the Part 2 (First Order ODEs)
Return to the Part 3 (Numerical Methods)
Return to the Part 4 (Second and Higher Order ODEs)
Return to the Part 5 (Series and Recurrences)
Return to the Part 6 (Laplace Transform)
Return to the Part 7 (Boundary Value Problems)